# Factoring degree 3 and higher polynomial not using factor theorem

I've been doing some googling and gandering over the inter webs and trying to find a method for factoring a cubic function without using the factor theorem (basically I want to know if there is a more efficient method of finding a linear factor), so does anyone know of a more efficient way of solving for a factor of a cubic? Basically what I'm expecting is too see different ways that it could be solved, much like how there are different ways of dividing polynomials, instead of doing guess and check.

• @Arthur Quick is dependent on how fast I can work that thing out and sometimes whether or not my calculator takes complex numbers – Simply Beautiful Art Oct 26 '16 at 17:05
• There is a cubic formula that I have never bothered to memorize. mathworld.wolfram.com/CubicFormula.html. I take it that you know the rational root theorem? It is very easy to synthetic division in an excel spreadsheet. And that is my first go-to if I have the resource available. – Doug M Oct 26 '16 at 17:14
• @DougM Seems he doesn't want anything along the lines of guess-and-check. – Simply Beautiful Art Oct 26 '16 at 17:20
• Most methods to factoring cubic polynomials take less than $5$ steps, though each step is rather lengthy. – Simply Beautiful Art Oct 26 '16 at 17:41

I'm going to post the method involving trig functions. As an example, we will be using $0=x^3-3x^2+3$.

To use this method, you must know that

$$\cos(3\theta)=4\cos^3(\theta)-3\cos(\theta)$$

which is easily derived from the sum of angles formulas and the Pythagorean theorem.

We manipulate this to give us

$$\cos(3\arccos(r))=4r^3-3r\tag0$$

$$0=ax^3+bx^2+cx+d$$

$$0=x^3-3x^2+3$$

We always start with the substitution $x=y-\frac{b}{3a}$, which for our example is $x=y+1$

$$0=(y+1)^3-3(y+1)^2+3\\=y^3-3y+1$$

In general, we get something along the lines of

$$0=a\left(y-\frac{b}{3a}\right)^3+b\left(y-\frac{b}{3a}\right)^2+c\left(y-\frac{b}{3a}\right)+d\\=ay^3+py+q$$

Where $p$ and $q$ are constants.

Make the substitution $y=uz$ and multiply both sides by $v$.

$$0=v(uz)^3-3v(uz)+3v\\=vu^3z^3-3vuz+v$$

And have $vu^3=4$ and $-3vu=-3$ so that it comes in our $(0)$ form. Solving this system of equations gives $u=2$ and $v=\frac12$.

$$0=4z^3-3z+\frac12$$

$$0=\cos(3\arccos(z))+\frac12$$

Solving for $z$ and recalling the period of cosine, we get

$$z=\cos\left(\frac{2\pi k}3+\frac13\arccos\left(\frac{-1}2\right)\right)$$

$$y=2z=2\cos\left(\frac{2\pi k}3+\frac13\arccos\left(\frac{-1}2\right)\right)$$

$$x=1+y=1+2\cos\left(\frac{2\pi k}3+\frac13\arccos\left(\frac{-1}2\right)\right)$$

$$x=1+2\cos\left(\frac{2\pi k}3+\frac{2\pi}9\right)\qquad k=1,2,3$$

This is not always possible, but under those cases, we can just switch to different trig functions and use their corresponding formulas. It's pretty easy to see which trig function you should use at the "solve for $u$ and $v$ step".

An advantage to this method is that it avoids casus irreducibilis, which is a fancy way of saying you can't factor a cubic polynomial using only real numbers, radicals, and basic arithmetic operations. This does not include trig functions, however, which can be used to obtain nice forms of the solution.

To compare, the radical solution for the above polynomial is given as

$$x_1=1-\frac12(1-i\sqrt3)\sqrt[3]{\frac12(-1+i\sqrt3)}-\frac{1+i\sqrt3}{\sqrt[3]{4(-1+i\sqrt3}}$$

$$x_2=1-\frac{1-i\sqrt3}{\sqrt[3]{4(-1+i\sqrt3)}}-\frac12(1+i\sqrt3)\sqrt[3]{\frac12(-1+i\sqrt3)}$$

$$x_3=1+\frac1{\sqrt[3]{\frac12(-1+i\sqrt3)}}+\sqrt[3]{\frac12(-1+i\sqrt3)}$$

Compared side by side, you might find one form much nicer.

• ...and these are why we don't teach our algebra students cubic polynomials... – Simply Beautiful Art Oct 26 '16 at 17:37